Problem: $\overline{AB}$ = $\sqrt{109}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $\sqrt{109}$ $?$ $ \sin( \angle BAC ) = \frac{10\sqrt{109} }{109}, \cos( \angle BAC ) = \frac{3\sqrt{109} }{109}, \tan( \angle BAC ) = \dfrac{10}{3}$
Answer: $\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{\sqrt{109}} $ $ \overline{BC}=\sqrt{109} \cdot \sin( \angle BAC ) = \sqrt{109} \cdot \frac{10\sqrt{109} }{109} = 10$